Mechanical Engineering Chapter 14 Homework Work And Energy Which Acts The Direction

397

14–21.

SOLUTION

Block A:

Block B:

Use the system of both blocks. NA,NB,T, and Rdo no work.

Substituting and solving,

+a©Fy=may;NA–60 cos 60° =0

Determine the velocity of the 60-lb block Aif the two

blocks are released from rest and the 40-lb block Bmoves

2 ft up the incline.The coefficient of kinetic friction

between both blocks and the inclined planes is mk=0.10.

A

B

398

14–22.

SOLUTION

Principle of Work and Energy: Here, the friction force

. Since the friction force is always opposite the motion,it does negative work.

Assume the block bounces back and stops without striking spring A.The spring

Since , the block stops before it strikes spring A.Therefore, the

s2=1.227 ft 62ft

10.0 lb

Ff=mkN=0.4(25) =

The 25-lb block has an initial speed of when it

is midway between springs Aand B. After striking spring B

it rebounds and slides across the horizontal plane toward

spring A, etc. If the coefficient of kinetic friction between

the plane and the block is determine the total

distance traveled by the block before it comes to rest.

mk=0.4,

v0=10 ft

>

s2ft

1ft

v0=10ft/s

kA=10 lb/in. kB=60lb/in.

BA

,

Ans:

399

14–23.

The 8-kg block is moving with an initial speed of 5 m

>

s. If

the coefficient of kinetic friction between the block and

plane is m

k=0.25,

determine the compression in the spring

when the block momentarily stops.

SOLUTION

Work. Consider the force equilibrium along y axis by referring to the FBD of the

block, Fig. a

Ans:

2 m

5 m/s

kA  200 N/m

A

B

400

*14–24.

SOLUTION

Kinematics: The speed of the block Aand Bcan be related by using the position

coordinate equation.

Equation of Motion:

Principle of Work and Energy: By considering the whole system, ,which acts

From Eq. [1], . Also,

¢sA=c(yA)0+yA

WA

At a given instant the 10-lb block Ais moving downward

with a speed of 6 ft s. Determine its speed 2 s later. Block B

has a weight of 4 lb, and the coefficient of kinetic friction

between it and the horizontal plane is . Neglect the

mass of the cord and pulleys.

mk=0.2

A

B

Ans:

401

14–25.

The 5-lb cylinder is falling from A with a speed vA

=10 ft>s

onto the platform. Determine the maximum displacement

of the platform, caused by the collision. The spring has an

unstretched length of 1.75 ft and is originally kept in

compression by the 1-ft long cables attached to the platform.

Neglect the mass of the platform and spring and any energy

lost during the collision.

SOLUTION

T1+Σ U1–2=T2

Ans:

A

3 ft

vA  10 ft/s

14–26.

SOLUTION

2sC+sA=l

The catapulting mechanism is used to propel the 10-kg

slider Ato the right along the smooth track. The propelling

action is obtained by drawing the pulley attached to rod BC

rapidly to the left by means of a piston P. If the piston

applies a constant force to rod BC such that it

moves it 0.2 m, determine the speed attained by the slider if

it was originally at rest. Neglect the mass of the pulleys,

cable, piston, and rod BC.

F=20 kN

FBC

A

P

Ans:

403

14–27.

SOLUTION

TA+©UA–B=TB

T

h

e “f

l

y

i

ng car”

i

s a r

id

e at an amusement par

k

w

hi

c

h

consists of a car having wheels that roll along a track

mounted inside a rotating drum. By design the car cannot

fall off the track, however motion of the car is developed by

applying the car’s brake, thereby gripping the car to the

track and allowing it to move with a constant speed of the

track, If the rider applies the brake when going

from Bto Aand then releases it at the top of the drum, A,

so that the car coasts freely down along the track to B

determine the speed of the car at Band the

normal reaction which the drum exerts on the car at B.

Neglect friction during the motion from Ato B.The rider

and car have a total mass of 250 kg and the center of mass of

the car and rider moves along a circular path having a

radius of 8 m.

1u=prad2,

vt=3m>s.

A

B

8m

v

t

u

Ans:

404

*14–28.

The 10-lb box falls off the conveyor belt at 5-ft

>

s. If the

coefficient of kinetic friction along AB is m

k=0.2,

determine the distance x when the box falls into the cart.

SOLUTION

Work. Consider the force equilibrium along the y axis by referring to Fig. a,

Principle of Work And Energy. Applying Eq. 14–7

Kinematics. Consider the vertical motion with reference to the x–y coordinate

system,

Solve for positive root,

Then, the horizontal motion gives

5 ft/s

30 ft

15 ft

B

y

4

35

A

Ans:

14–29.

SOLUTION

T1+©U1–2=T2

The collar has a mass of 20 kg and slides along the smooth

rod.Two springs are attached to it and the ends of the rod as

shown. If each spring has an uncompressed length of 1 m

and the collar has a speed of when determine

the maximum compression of each spring due to the back-

and-forth (oscillating) motion of the collar.

s=0,2m>s

s

0.25 m

1m 1m

k

A

50 N/mk

B

100 N/m

Ans:

406

14–30.

The 30-lb box Ais released from rest and slides down along

the smooth ramp and onto the surface of a cart. If the cart

is prevented from moving, determine the distance sfrom the

end of the cart to where the box stops.The coefficient of

kinetic friction between the cart and the box is mk=0.6.

SOLUTION

Principle of Work and Energy: which acts in the direction of the vertical

WA

10 ft

4ft

s

A

BC

Ans:

407

14–31.

SOLUTION

TA+©

UA–B=TB

2m

3m

A

B

C

Marbles having a mass of 5 g are dropped from rest at A

through the smooth glass tube and accumulate in the can

at

C.

Determine

the

placement

R

of

the

can

from

the

end

of the tube and the speed at which the marbles fall into the

can. Neglect the size of the can.

Ans:

408

*14–32.

SOLUTION

lAC =2(0.3)2+(0.4)2=0.5 m

T

h

e

bl

oc

k

h

as a mass of 0.8

k

g an

d

moves w

i

t

hi

n t

h

e smoot

h

vertical slot. If it starts from rest when the attached spring is

in the unstretched position at A, determine the constant

vertical force Fwhich must be applied to the cord so that

the block attains a speed when it reaches B;

Neglect the size and mass of the pulley. Hint:

The work of Fcan be determined by finding the difference

in cord lengths AC and BC and using U

F=F¢l.¢l

sB=0.15 m.

vB=2.5 m>s

B

C

A

0.4 m

0.3 m

s

B

409

14–33.

SOLUTION

Solving for the positive root,

TA+©UA–B=TB

The 10-lb block is pressed against the spring so as to

compress it 2 ft when it is at A. If the plane is smooth,

determine the distance d, measured from the wall, to where

the block strikes the ground. Neglect the size of the block.

3ft

4ftd

B

A

k100 lb/ft

Ans:

410

14–34.

SOLUTION

The spring bumper is used to arrest the motion of the

4-lb block, which is sliding toward it at As

shown, the spring is confined by the plate Pand wall using

cables so that its length is 1.5 ft. If the stiffness of the

spring is determine the required unstretched

length of the spring so that the plate is not displaced more

than 0.2 ft after the block collides into it. Neglect friction,

the mass of the plate and spring,and the energy loss

between the plate and block during the collision.

k=50 lb>ft,

v=9ft>s. P

k

v

A

1.5 ft. 5 ft.

A

Ans:

411

14–35.

SOLUTION

y=50 cos ap

100 bx

`

x=35

=22.70 ft

When the 150-lb skier is at point Ahe has a speed of

Determine his speed when he reaches point Bon the

smooth slope.For this distance the slope follows the cosine

curve shown. Also, what is the normal force on his skis at B

and his rate of increase in speed? Neglect friction and air

resistance.

5ft

>

s.

A

B

y

y50 cos ( x)

100

p

Ans:

412

*14–36.

SOLUTION

5b=0

The spring has a stiffness and an unstretched

length of2ft. As shown, it is confined by the plate and wall

using cables so that its length is 1.5 ft. A 4-lb block is given a

speed when it is at A,and it slides down the incline having

acoefficient of kinetic friction If it strikes the plate

and pushes it forward 0.25 ft before stopping,determine its

speed at A.Neglect the mass of the plate and spring.

mk=0.2.

vA

k=50 lb>ft

3ft

1.5 ft

A

k50 lb/ft

3

4

5

v

A

Ans:

413

14–37.

If the track is to be designed so that the passengers of the

roller coaster do not experience a normal force equal to

zero or more than 4 times their weight, determine the

limiting heights and so that this does not occur.The

roller coaster starts from rest at position A. Neglect friction.

hC

hA

SOLUTION

Equations of Motion: Here,.The requirement at position Bis that

. By referring to Fig. a,

Principle of Work and Energy: The normal reaction Ndoes no work since it always

We have

When the rollercoaster moves from position Ato C,Wdisplaces vertically

downward

h=hA–hC=(22.5 –hC)m.

NB=4mg

an=v2

r

A

hA

C

B

hC

rC20 m

rB15 m

Ans:

414

14–38.

SOLUTION

Principle of Work and Energy: By referring to Fig. a,we notice that Ndoes no work since

If the 60-kg skier passes point Awith a speed of ,

determine his speed when he reaches point B. Also find the

normal force exerted on him by the slope at this point.

Neglect friction.

5m>s

y

B

A

15 m

y(0.025x25) m

Ans:

14–39.

If the 75-kg crate starts from rest at A, determine its speed

when it reaches point B.The cable is subjected to a constant

force of . Neglect friction and the size of the

pulley.

F=300 N

B

C

A

6m 2 m

6m

30

F

SOLUTION

Principle of Work and Energy: By referring to Fig. a, notice that N,W, and Rdo no

Ans:

416

*14–40.

If the 75-kg crate starts from rest at A, and its speed is

when it passes point B, determine the constant force F

exerted on the cable. Neglect friction and the size of the

pulley.

6m>s

B

C

A

6m 2 m

6m

30

F

SOLUTION

Free-Body Diagram: The free-body diagram of the crate and cable system at an

Ans: